UVA 10655 - Contemplation! Algebra

题目链接

题意：给定p, q, n代表p=a+b,q=ab求an+bn

思路：矩阵快速幂，公式变换一下得到(an+bn)(a+b)=an+1+bn+1+ab(an?1+bn?1)，移项一下得到an+1+bn+1=(an+bn)p?q(an?1+bn?1)

这样就可以用矩阵快速幂求解了

代码：

#include <stdio.h>
#include <string.h>

long long p, q, n;

struct mat {
	long long v[2][2];
	mat() {memset(v, 0, sizeof(v));}
	mat operator * (mat c) {
		mat ans;
		for (int i = 0; i < 2; i++) {
			for (int j = 0; j < 2; j++) {
				for (int k = 0; k < 2; k++) {
					ans.v[i][j] = ans.v[i][j] + v[i][k] * c.v[k][j];
    			}
   			}
  		}
  		return ans;
 	}
};

mat pow_mod(mat x, long long k) {
	mat ans;
	ans.v[0][0] = ans.v[1][1] = 1;
	while (k) {
		if (k&1) ans = ans * x;
		x = x * x;
		k >>= 1;
 	}
 	return ans;
}

long long solve() {
	if (n == 0) return 2;
	if (n == 1) return p;
	if (n == 2) return p * p - 2 * q;
	mat a;
	a.v[0][0] = p; a.v[0][1] = -q; a.v[1][0] = 1;
	a = pow_mod(a, n - 2);
	return a.v[0][1] * p + a.v[0][0] * (p * p - 2 * q);
}

int main() {
	while (scanf("%lld%lld%lld", &p, &q, &n) == 3) {
		printf("%lld\n", solve());
	}
	return 0;
}